Prove that if $\lim_{n\to\infty}\frac{x_{n+1}}{x_n} = L < 1$ then $\lim_{n\to\infty} x_n = 0$ Let $(x_n)$ be a sequence with $x_n > 0$ for all $n \in \mathbb{N}$. I would like a hint on how to prove that if $\lim_{n\to\infty}\frac{x_{n+1}}{x_n} = L < 1$ then $\lim_{n\to\infty} x_n = 0$. 
Here is what i have tried. There exist a $N$ such that $|\frac{x_{n+1}}{x_n} - L| < \varepsilon$ for all $n \geq N$. From this i must conclude that for some $M$ we have $|x_n|<\varepsilon$ for $n \geq M$. Can anyone point me in the right direction? I suspect i need some smart trick. 
Thanks.
 A: How about this reverse argument: since $\lim_{n\to \infty}x_{n+1}/x_n=L<1$, then, by the ratio test, the series
$$
\sum_{n=1}^\infty x_n < \infty
$$
That gives 
$$
\lim_{n\to \infty}x_n=0
$$
A: Suppose the limit exists, and is different than $0$. Let's call it $x$. Then, $\lim \frac{x_{n+1}}{x_n}=\frac{x}{x}=1<1$. We have arrived at a contradition.
Therefore, if it exists, it is $0$. 
Pick $L<\gamma<1$. Therefore, there exists $N$ such that for all $n>N$, $x_{n+1}<\gamma x_n<x_n$. We then have the sequence is decreasing after $N$. But if it is decreasing and bounded, it converges.
We can summarize the argument as follows:

Okay, let's suppose the limit exists. Then it must be $0$. But the sequence is eventually monotonic and bounded. Therefore, it exists, and is $0$.

A: Since $L<1$ so choose $\epsilon$ such that $L+\epsilon, L-\epsilon <1$. Now applying the definition of limit to this $\epsilon$ and then applying recursion, you can see that $(L-\epsilon)^mx_N < x_{N+m}<(L+\epsilon)^m$. Now use the Sandwich Theorem.
A: If $\lim_{n\to\infty}\frac{x_{n+1}}{x_n}=L$, then for some $N$, $0<\frac{x_{n+1}}{x_n}<q\forall n>N$ where $q$ is some number between $L$ and $1$. (You can set $q$ as $L+\varepsilon$ where $\epsilon$ is small enough so that $L+\varepsilon<1$).
Now, notice that for all $n>N$, $0<x_{N+p}=\frac{x_{N+p}}{x_{N+p-1}}\cdot\frac{x_{N+p-1}}{x_{N+p-2}}\cdot\ldots\cdot\frac{x_{N+1}}{x_N}\cdot c<c\cdot q^{p}$
Where $c$ is a constant equal to $\frac{x_N}{x_{N-1}}\cdot\ldots\cdot\frac{x_2}{x_1}$
Taking the limit of the right side and noting that $|q|<1$, we find that the limit of the sequence $x_n$ is equal to $0$. 
QED.
A: Use inequality, 2nd form:
$$\Biggl\lvert\frac{x_{n+1}}{x_n}-L\Biggr\rvert\le\Biggl\lvert\frac{x_{n+1}}{x_n}-L\Biggr\rvert$$
If the latter is $<\varepsilon$ for all $n\ge N$, you deduce that:
$$0<\frac{x_{n+1}}{x_n}<L+\varepsilon,$$
hence if $L+\varepsilon\le k <1$, we have  $\,0<x_n<\dfrac{k^N}{x_N}k^{n}$,   i. e. if  $n$ is large enough , $x_n$ is bounded from above by a geometric sequence that converges to $0$.
A: We can find an $N$ so that if $n\ge N$, we have $\frac{x_{n+1}}{x_n}\le\frac{L+1}2$. This means that for $n\ge N$, $x_n$ is a decreasing sequence bounded below by $0$. Therefore, $\lim\limits_{n\to\infty}x_n$ exists and is $\ge0$.
Suppose that
$$
\lim_{n\to\infty}x_n=x_\infty
$$
Taking the limit of
$$
x_{n+1}\le\frac{L+1}2x_n
$$
yields
$$
x_\infty\le\frac{L+1}2x_\infty
$$
Subtracting the right side from both sides and multiplying by $\frac2{1-L}$, which is positive, gives
$$
x_\infty\le0
$$
Therefore, $x_\infty=0$.
